1/ |2x-8|=x+2 2/ |x-3|=|2x-1| 3/ -2x^2+|x+2|-2>=0 4/|2x-1| 6x-3

0 bình luận về “1/ |2x-8|=x+2 2/ |x-3|=|2x-1| 3/ -2x^2+|x+2|-2>=0 4/|2x-1|<x+2 5/ Căn(8+2x-x^2) > 6x-3”

1. Đáp án:

    5) \(x \in \left[ {\dfrac{1}{2};1} \right)\)

   Giải thích các bước giải:

   \(\begin{array}{l}
   1)\left| {2x – 8} \right| = x + 2\\
    \to \left[ \begin{array}{l}
   2x – 8 = x + 2\\
   2x – 8 =  – x – 2
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   x = 10\\
   3x = 6
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   x = 10\\
   x = 2
   \end{array} \right.\\
   2)\left| {x – 3} \right| = \left| {2x – 1} \right|\\
    \to \left[ \begin{array}{l}
   x – 3 = 2x – 1\\
   x – 3 =  – 2x + 1
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   x =  – 2\\
   3x = 4
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   x =  – 2\\
   x = \dfrac{4}{3}
   \end{array} \right.\\
   3)\left| {x + 2} \right| \ge 2{x^2} + 2\\
    \to \left[ \begin{array}{l}
   x + 2 \ge 2{x^2} + 2\\
   x + 2 \le  – 2{x^2} – 2
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   2{x^2} – x \le 0\\
   2{x^2} + x + 4 \le 0\left( {vô lý} \right)
   \end{array} \right.\\
    \to x \in \left[ {0;\dfrac{1}{2}} \right]\\
   4)\left| {2x – 1} \right| < x + 2\\
    \to 4{x^2} – 4x + 1 < {x^2} + 4x + 4\\
    \to 3{x^2} – 8x – 3 < 0\\
    \to x \in \left( { – \dfrac{1}{3};3} \right)\\
   5)\sqrt {8 + 2x – {x^2}}  > 6x – 3\\
    \to \left\{ \begin{array}{l}
   6x – 3 \ge 0\\
   8 + 2x – {x^2} > 36{x^2} – 36x + 9
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   x \ge \dfrac{1}{2}\\
   x \in \left( {\dfrac{1}{{37}};1} \right)
   \end{array} \right.\\
    \to x \in \left[ {\dfrac{1}{2};1} \right)
   \end{array}\)

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